MX2 Marathon (2 Viewers)

HazzRat · Dec 19, 2023

Question:
How did they get from this step to the other step shown below? I feel like i'm missing smthn pretty obvious

xoNat · Dec 19, 2023

HazzRat said:
Question:
How did they get from this step to the other step shown below? I feel like i'm missing smthn pretty obvious
View attachment 41860

Compound angle formulas, they're on the reference sheet

HazzRat · Dec 20, 2023

Can someone smarter than me plz tell me how they did these two steps? If it helps it's for this question:

Working out

scaryshark09 · Dec 20, 2023

for the first one, i think its meant to be x^2, instead of 2^x

all they do is split up the 2x into x+x and then they rearrange the algebra for the first step

HazzRat · Dec 20, 2023

scaryshark09 said:
for the first one, i think its meant to be x^2, instead of 2^x

all they do is split up the 2x into x+x and then they rearrange the algebra for the first step

Ahh makes sense. I was wondering abt that.

scaryshark09 · Dec 20, 2023

for the second one they are just factorising, but i think its meant to be xm instead of just x

scaryshark09 · Dec 20, 2023

scaryshark09 said:
for the second one they are just factorising, but i think its meant to be xm instead of just x

and again its meant to be x^2, not 2^x

HazzRat · Dec 20, 2023

scaryshark09 said:
for the second one they are just factorising, but i think its meant to be xm instead of just x

Fair enough. Just some bad working out answers ig

Luukas.2 · Dec 20, 2023

Yes, applying the corrections that have been noted above by @scaryshark09, the working should be:

$\begin{align*} \text{LHS} &= x\left(x^{k+2}\right) + (x + 1)^2(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + 2x + 1\right)(x + 1)^{2k + 1} \\ &= x\left(x^{k+2}\right) + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} + x(x + 1)^{2k + 1} \\ &= x\left[x^{k+2} + (x + 1)^{2k + 1}\right] + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} \\ &= x\left[m\left(x^2 + x + 1\right)\right] + \left(x^2 + x + 1\right)(x + 1)^{2k + 1} \qquad \text{using the induction hypothesis $P(k)$} \\ &= \left(x^2 + x + 1\right)\left[mx + (x + 1)^{2k + 1}\right] \\ &= p\left(x^2 + x + 1\right) \qquad \text{where $p = mx + (x + 1)^{2k + 1} \in \mathbb{Z}$ as $m$, $x$, $k \in \mathbb{Z}$} \\ &= \text{RHS} \end{align*}$

HazzRat · Jan 16, 2024

Have I been smoking crack or what is this

carrotsss · Jan 16, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

e^(-pi/2i) is just 1cis(-pi/2) which is -i, then multiplying that by -2i is pretty simple

Average Boreduser · Jan 16, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

convert e^...sdjkjfhwkjehgkle to cis form and the i is basically xing cispi/2?

Luukas.2 · Jan 17, 2024

HazzRat said:
Have I been smoking crack or what is this View attachment 42150

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= -2 \times 1\left(e^{\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|i| = 1$ and $\arg{i} = \frac{\pi}{2}$} \\ &= -2\left(e^0\right) \\ &= -2 \times 1 \\ &= -2 \end{align*}$

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= 2 \times 1\left(e^{-\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|-i| = 1$ and $\arg{(-i)} = -\frac{\pi}{2}$} \\ &= 2\left(e^{-\frac{\pi}{2}i}\right)^2 \\ &= 2\left(e^{-i\pi}\right)\\ &= \frac{2}{e^{i\pi}} \\ &= \frac{2}{-1} \qquad \text{as $e^{i\pi} = -1$ is the Euler identity} \\ &= -2 \end{align*}$

HazzRat · Jan 17, 2024

Luukas.2 said:
Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= -2 \times 1\left(e^{\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|i| = 1$ and $\arg{i} = \frac{\pi}{2}$} \\ &= -2\left(e^0\right) \\ &= -2 \times 1 \\ &= -2 \end{align*}$

Or...

$\begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= 2 \times 1\left(e^{-\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|-i| = 1$ and $\arg{(-i)} = -\frac{\pi}{2}$} \\ &= 2\left(e^{-\frac{\pi}{2}i}\right)^2 \\ &= 2\left(e^{-i\pi}\right)\\ &= \frac{2}{e^{i\pi}} \\ &= \frac{2}{-1} \qquad \text{as $e^{i\pi} = -1$ is the Euler identity} \\ &= -2 \end{align*}$

In the end it was just cuz I didn't know eulers formula

Luukas.2 · Jan 17, 2024

HazzRat said:
In the end it was just cuz I didn't know eulers formula

Cool... I was just showing some other approaches

HazzRat · Jan 18, 2024

Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

Average Boreduser · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

just look up properties lol. no way of getting around that

liamkk112 · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

u just got to memorise the quadrilateral properties no way around it

usually though:
- parallelogram -> pairs of equal side lengths, parallel sides
- square -> equal side lengths, 90 degrees between sides, parallel sides
- rectangle -> 90 degrees between sides, parallel sides
- rhombus -> pairs of equal side lengths, parallel sides, diagonals meet at 90 degrees and bisect

there r also kites but i forget how those work and they're relatively uncommon

Luukas.2 · Jan 18, 2024

HazzRat said:
Does anyone have a cheat sheet for proving shapes in complex numbers? Whenever I'm given a question like "prove these complex points form a parallelogram" I never know how to prove it and the answer's always smthn random like "the diagonals bisect each other". So is there a method of knowing the proof for each shape?

There is always a purely algebraic method, which is usually awful. There are sometimes purely geometric methods (like for arg(z - i) = arg(z + 1) etc.). If there isn't an obvious purely geometric approach, the efficient answer is likely to involve:

treating the complex numbers as vectors
looking for geometric properties that proves the required result
demonstrating these properties through algebraic representation of vectors

For example... the complex number z represents a point A in the first quadrant. If O is the origin, B lies in the second quadrant, and OACB is a square, find the complex number representing point C. Under what conditions is C located in the second quadrant.

A diagram should make it obvious that side OB is adjacent to side OA in the square.

Properties of a square then dictate that OB = i.OA, and so the complex number iz represents B.

Then, using vector reasoning:

$\begin{align*} \overrightarrow{OC} &= \overrightarrow{OA} + \overrightarrow{AC} \\ &= \overrightarrow{OA} + \overrightarrow{AB} \qquad \text{as $\overrightarrow{AC} = \overrightarrow{OB}$ as opposite sides of a square} \\ &= z + iz \\ &= z(1 + i) \end{align*}$

Hence, the point C is represented by z(1 + i), and so is in the second quadrant if

$\frac{\pi}{4} < \arg{z} < \frac{\pi}{2}$

from the diagram (as A must be in quadrant 1 and, for C to be in quadrant 2 given angle COA is 45 degrees, OA must be inclined at at least 45 degrees above the real axis), or (algebraically), by solving:

$\begin{align*} \frac{\pi}{2} &< \arg{[z(1 + i)]} < \pi \\ \frac{\pi}{2} &< \arg{z} + \arg{(1 + i)} < \pi \\ \frac{\pi}{2} &< \arg{z} + \frac{\pi}{4} < \pi \\ \frac{\pi}{4} &< \arg{z} < \frac{3\pi}{4} \qquad \qquad \text{. . . (1)} \\ \\ \text{But, as $A$ lies in quadrant 1:} \qquad 0 &< \arg{z} < \frac{\pi}{2} \qquad \qquad \text{. . . (2)} \\ \\ \text{Thus, to satisfy both (1) and (2):} \qquad \frac{\pi}{4} &< \arg{z} < \frac{\pi}{2} \end{align*}$

HazzRat · Jan 24, 2024

I

this question so much

(doesn't need solving. I was a big boy and did it myself)

Make a Difference – Donate to Bored of Studies!

MX2 Marathon (2 Viewers)

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

don't worry, be happy

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

∞∆ who let 'em cook dis long ∆∞

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

∞∆ who let 'em cook dis long ∆∞

∞∆ who let 'em cook dis long ∆∞

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Well-Known Member

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

New Member

Rising Renewal

Well-Known Member

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Well-Known Member

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Rising Renewal

Well-Known Member

Well-Known Member

H̊ͯaͤz͠z̬̼iẻͩ̊͏̖͈̪

Users Who Are Viewing This Thread (Users: 0, Guests: 2)